已知数列{an},等差中项为四分之五
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![已知数列{an},等差中项为四分之五](/uploads/image/f/4267518-6-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%2C%E7%AD%89%E5%B7%AE%E4%B8%AD%E9%A1%B9%E4%B8%BA%E5%9B%9B%E5%88%86%E4%B9%8B%E4%BA%94)
由题意,2sn=[(an+2)/2]的平方,sn=an平方/8+an/2+1/2,则s(n-1)=a(n-1)平方+a(n-1)/2+1/2,两式相减得:sn-s(n-1)=an=(an平方-an-1
1、由题意,得(a1+2)/2=√(2a1)整理,得(a1-2)²=0a1-2=0a1=2(an+2)/2=√(2Sn)整理,得8Sn=(an+2)²8Sn-1=[a(n-1)+2
第一步的前提是n≥2,所以A(n+1)=3An只对n≥2有效所以还是要求出a2才能求a5
由题意知对任意n有2S[n]=a[n]^2+a[n]同样有:2S[n-1]=a[n-1]^1+a[n-1]两式相减,得左边=2S[n]-2S[n-1]=2a[n]即2a[n]=a[n]^2+a[n]-
看图片:前三项2,6,10(2)由题意,2sn=[(an+2)/2]的平方,sn=an平方/8+an/2+1/2,则s(n-1)=a(n-1)平方+a(n-1)/2+1/2,两式相减得:sn-s(n-
①依题意,得(an+2)/2=根号下(2Sn),∴a1+2=2根号下(2S1)=2根号下(2a1),∴(a1-2)的平方=0,∴a1=2,由a2+2=2根号下(2S2)=2根号下[2(2+a2)],得
1)由题意得,a1=1,当n>1时,sn=an^2/2+an/2sn-1=a(n-1)^2/2+a(n-1)/2,∴sn-sn-1=an^2/2-a(n-1)^2/2+an/2-a(n-1)/2即(a
(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
(1)由题知,Sn-1是an与-3的等差中项.∴2Sn-1=an-3即an=2Sn-1+3(n≥2,n∈N*)…(2分)a2=2S1+3=2a1+3=9a3=2S2+3=2(a1+a2)+3=27a4
设等比数列{an}的公比为q,则:a2=a1q,a3=a1q2,由a3是a1,a2的等差中项,得:2a3=a1+a2,即2a1q2=a1+a1q,因为a1≠0,所以2q2-q-1=0,解得:q=−12
(1)an是Sn与2的等差中项即a1=2sn=2an-2所以s(n-1)=2a(n-1)-2an=sn-s(n-1)=2a(n-1)所以an为等比数列公比为2首项为2则an=2^n而点P(bn,bn+
an是n与Sn的等差中项,即:an-n=Sn-an,亦即:2an=n+Sn令n=1,代入得a1=1当n≥2时:2an=n+Sn;2a(n-1)=(n-1)+S(n-1)二式相减:2an-2a(n-1)
解2an=n+SnSn=2an-n(1)S(n-1)=2a(n-1)-n+1做差的an=2an-2a(n-1)+1an=2a(n-1)+1an+1=2[a(n-1)+1]即[an+1]/[a(n-1)
(1)2an=n+Sn2a(n+1)=n+1+S(n+1)相减得2【a(n+1)-an】=1+a(n+1)a(n+1)=2an+1b(n+1)=a(n+1)+1=2(an+1)=2bna1=1an=2
因:Sn是An和1的等差中项所以有:2Sn=An+1即:Sn=(An+1)/2An=Sn-S(n-1)=(An+1)/2-[A(n-1)+1]/2=[An-A(n-1)]/2An=-A(n-1)A1=
Sn与2的等比中项为√(2Sn),an与2的等差中项为(an+2)/2由题目可知,8Sn=(an+2)^2,所以8S_(n-1)=[a_(n-1)+2]^2.两者相减,得8an=an^2+4an-[a
因为a1a5=20,a2+a4=-12{a}等比,所以a2a4=20,a2,a4是方程:x^2+12x+20=0的根x=-2或x=-101.a2=-2,a4=-10q^2=a4/a2=5a8=a4*q
(a3+1)是a2,a3的等差中项2(a3+1)=a2+a3a3-a2=-2数列递减与已知好像矛盾再问:已知递增等比数列{an}满足a2a3a4=64,且(a3+1)是a2,a3的等差中项,求数列{a
由题意得(an+1)/2=√(Sn×1)Sn=[(an+1)/2]²n=1时,S1=a1=[(a1+1)/2]²,整理,得(a1-1)²=0a1=1n≥2时,Sn=[(a